1 Excel functions =NORMSDIST( z value) =NORMSINV ( probability ...

convenience, there is also a column recording whether a student lost weight (0 = “no weight
lost”, 1 = “some weight lost”).
The pulled records reveal that only 24 on the 40 graduates achieved any weight loss during
the month. That means that only 60% of the records pulled demonstrate any weight loss. The
question you must now consider is: is this evidence strong enough to justify an accusation of
false advertising by Beautiful U?
THE QUESTION:
Suppose that the average weight lost by a student in the program is 5 pounds, with a
standard deviation of 20 pounds. How likely is it that the average weight loss in a
random sample of 40 students will be 8.45 pounds or more?
4. Our task to find the probability that something is greater than or equal to 8.45. What is that “something”?
a) z b) σ c) s d) µ e) x (x-bar)
5. Our sample size is 40, so we’ll be able to apply our 191 techniques to answer THE QUESTION. This is
because the histogram of a 40 student sample suggests that the population is
a) perfectly normal b) roughly normal c) roughly symmetric d) roughly binomial e) bimodal
6. Question #2 says that “we’ll be able to apply our 191 techniques to answer THE QUESTION”. This is
because the observation that we made in Question #2 allows us to conclude that
a). the population is approximately normal
b). µ is approximately normal
c). the sampling distribution of the mean is approximately normal
d). s is a good approximation for σ
e). µ is a good approximation for x .
7. To answer THE QUESTION, we need to find
the standard deviation of the sampling
distribution of the mean. It is equal to
a). 20
b). 20/40 = 0.5
c). (8.45 - 5)/20 = 0.1725
d). SQRT(8.45 × (20 – 8.45)/40) = 1.562
e). 20/SQRT(40) = 3.162
2
8. We also need to know the mean of the
sampling distribution of the mean. In this
problem, this is
a). =NORMSINV(0.1725) = -0.9443
b). =NORMSINV(1-0.1725) = 0.9443
c). 5
d). =SQRT(40) = 6.325
e). 8.45
9. Which calculation would give the probability that one student chosen at random from the Beautiful U
graduates would have lost 8.45 pounds or more? (Assume, for this problem only, that weight loss is
normally distributed.)
a). = NORMDIST( 5, 8.45, 20, TRUE)
b). = NORMDIST( 5, 8.45, 3.162, TRUE)
c). = NORMDIST( 8.45, 5, 20, TRUE)
d). =1 - NORMDIST( 8.45, 5, 20, TRUE)
e). =1 - NORMDIST( 8.45, 5, 3.162, TRUE)
10. Suppose that the answer to THE QUESTION is 0.1376. What does this mean?
a). About 14% of all Beautiful U students lose 5 pounds or more.
b). About 14% of all Beautiful U students lose 8.45 pounds or more.
c). About 14% of the students in a sample of 40 students would be expected to lose 5 pounds or more.
d). About 14% of the students in a sample of 40 students would be expected to lose 8.45 pounds or more.
e). None of these interpretations is correct.